September 13th, 2009, 05:54 PM  #1 
Newbie Joined: Sep 2009 Posts: 1 Thanks: 0  how can these to be factor?
3x^2  7x + 3 and 2x^3 + 128 i have no idea how the first one can be factor... i dont think it can be factor... the second one, i went: 2[(x^3 + 64)] could this be answer? seems like i could go more steps 
September 14th, 2009, 06:02 AM  #2 
Senior Member Joined: Mar 2009 Posts: 318 Thanks: 0 
To learn how to factor quadratics (something they were supposed to have covered back in early algebra), try here. For the second one, you need now to apply the differenceofcubes factoring formula. (Memorize this, along with the sumofcubes formula and the differenceofsquares formula. You will need all of these!) 
September 14th, 2009, 10:25 PM  #3 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
We can possibly factor 3x²  7x + 3 using irrational numbers. Let the given expression equal to 0, so 3x²  7x + 3 = 0. Using quadratic formula for this equation, we arrive with roots of x = (7 ± ?(13)) / 6. Since 3x²  7x + 3 = 0 is equivalent to x²  (7/3)x + 1 = 0, it's clear to see that (x  (7 + ?(13)) / 6)(x  (7  ?(13)) / 6) = 0, then it's obvious that 3(x  (7 + ?(13)) / 6)(x  (7  ?(13)) / 6) = 3x²  7x + 3 = 0. ? 3(x  (7 + ?(13)) / 6)(x  (7  ?(13)) / 6). To factor the expression 2x³ + 128 (or 2(x³ + 4³)), use the suggested link given by stapel, we have 2(x + 4)(x²  4x + 16). 

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